1.A bus travels 400 m between two stops. It starts from rest and accelerates at 1.50 m/s^2 until it reaches a velocity of 9.00 m/s. The bus continues at this velocity and then decelerates at 2 m/s^2 until it comes to a halt. Find the total time required for the journey.

2.A 28-kg block is connected to an empty 1kg bucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.45 and the coefficient of kinetic friction between the table and block is 0.32. Sand is gradually added to the bucket until the system just begins to move.
a)Calculate the amount of sand added to the bucket.
b)calculate the acceleration of the system.

How did the 10.1 come about after the solving of 9/2*4.5?

1. Ah, the bus journey from start to stop, quite the adventure! Let's calculate the total time required for this bumpy ride.

First, let's find the time it takes for the bus to accelerate to its maximum velocity of 9.00 m/s. We can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Rearranging the formula, we have t = (v - u) / a.

Using the given values, the initial velocity (u) is 0 m/s, the final velocity (v) is 9.00 m/s, and the acceleration (a) is 1.50 m/s^2. Plugging these values into the formula, we get t = (9.00 - 0) / 1.50 = 6 seconds.

Next, the bus continues at a constant velocity of 9.00 m/s until it starts decelerating. Since this is a constant velocity, the time it takes for this part of the journey is simply the distance divided by the velocity. Given that the distance is 400 m and the velocity is 9.00 m/s, the time is t = distance / velocity = 400 / 9.00 = 44.44 seconds (approximately).

Finally, the bus decelerates at 2 m/s^2 until it comes to a halt. We can once again use the formula v = u + at and solve for t. Since the final velocity (v) is 0 m/s and the deceleration (a) is 2 m/s^2, we have t = (0 - 9.00) / -2 = 4.5 seconds.

To find the total time required for the journey, we add up the times for each segment: 6 seconds + 44.44 seconds + 4.5 seconds = 54.94 seconds (approximately).

So, the total time required for this hilarious bus journey is approximately 54.94 seconds. And don't worry, Clown Bot is not the driver!

2. Ah, the classic tale of a block and a bucket trying to get their sand-filled groove on! Let's calculate the amount of sand added and the acceleration of this slippery system.

a) The system will just begin to move when the static friction between the table and the block is overcome. We need to find the maximum amount of sand that can be added to the bucket before this happens. The maximum force of static friction can be calculated using the formula F_max_static = μ_static * N, where μ_static is the coefficient of static friction and N is the normal force.

In this case, N is the force due to gravity acting on the block. We can calculate it using the formula N = m_block * g, where m_block is the mass of the block and g is the acceleration due to gravity.

Given that the mass of the block (m_block) is 28 kg and the coefficient of static friction (μ_static) is 0.45, we have F_max_static = 0.45 * (28 kg * 9.8 m/s²) ≈ 123.48 N.

Since the bucket is empty initially, there is no sand adding any extra force. Therefore, the maximum force of static friction (F_max_static) must support the weight of both the block and the bucket. We can consider the system as having a total mass of 29 kg (28 kg block + 1 kg bucket). So, we have F_max_static = (29 kg * 9.8 m/s²) = 284.2 N.

To determine the amount of sand added to the bucket, the maximum force of static friction must equal the force due to gravity acting on the block and the sand. So, we can use the formula F_sand = m_sand * g, where F_sand is the force due to gravity acting on the added sand, m_sand is the mass of the sand, and g is the acceleration due to gravity.

Plugging in the values we found for F_max_static (284.2 N) and g (9.8 m/s²), we have 284.2 N = m_sand * 9.8 m/s². Solving for m_sand, we find that m_sand ≈ 29.02 kg.

Therefore, the amount of sand added to the bucket is approximately 29.02 kg.

b) Now let's determine the acceleration of the system once the sand is added. To calculate this, we need to consider the forces acting on the system. The force of kinetic friction between the table and the block will oppose the motion of the system.

The force of kinetic friction (F_kinetic) can be calculated using the formula F_kinetic = μ_kinetic * N, where μ_kinetic is the coefficient of kinetic friction and N is the normal force (which we already found to be 29 kg * 9.8 m/s²).

Given that the coefficient of kinetic friction (μ_kinetic) is 0.32, we have F_kinetic = 0.32 * (29 kg * 9.8 m/s²) ≈ 89.04 N.

Since the force of kinetic friction acts in the opposite direction to the force due to gravity on the system, the net force on the system is F_net = F_sand - F_kinetic.

Plugging in the values we found for F_sand (approximately 284.2 N) and F_kinetic (approximately 89.04 N), we have F_net = 284.2 N - 89.04 N = 195.16 N.

Finally, we can use Newton's second law, F_net = m_total * a, where F_net is the net force, m_total is the total mass of the system (29 kg), and a is the acceleration. Solving for a, we find a = F_net / m_total = 195.16 N / 29 kg ≈ 6.73 m/s².

Therefore, the acceleration of the system once the sand is added is approximately 6.73 m/s². Sand is a real mover and shaker!

1. To find the total time required for the journey of the bus, we need to calculate the time taken for each phase of the motion - acceleration, constant velocity, and deceleration.

The first phase is acceleration. We can use the equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Here, the final velocity (v) is 9.00 m/s, the initial velocity (u) is 0 m/s (as the bus starts from rest), and the acceleration (a) is 1.50 m/s^2.

Using the equation, we can rearrange it as t = (v - u) / a and calculate the time taken for the acceleration phase.

t1 = (9.00 m/s - 0 m/s) / 1.50 m/s^2
t1 = 6 seconds

So, the time taken for the acceleration phase is 6 seconds.

The second phase is the constant velocity phase, where the bus continues to move at a constant velocity of 9.00 m/s. The time taken for this phase can be calculated using the equation, distance = velocity × time.

The distance traveled during this phase is 400 m, and the velocity is 9.00 m/s.

t2 = distance / velocity
t2 = 400 m / 9.00 m/s
t2 = 44.44 seconds (rounded to two decimal places)

So, the time taken for the constant velocity phase is approximately 44.44 seconds.

The third phase is deceleration. We can use the same equation of motion, v = u + at, but the acceleration will be negative as the bus is decelerating.

The final velocity (v) is 0 m/s (as the bus comes to a halt), the initial velocity (u) is 9.00 m/s, and the acceleration (a) is -2.00 m/s^2.

t3 = (0 m/s - 9.00 m/s) / -2.00 m/s^2
t3 = 4.5 seconds

So, the time taken for the deceleration phase is 4.5 seconds.

The total time required for the journey is the sum of the time taken for each phase.

Total time = t1 + t2 + t3
Total time = 6 seconds + 44.44 seconds + 4.5 seconds
Total time = 54.94 seconds (rounded to two decimal places)

Therefore, the total time required for the journey of the bus is approximately 54.94 seconds.

2. a) To calculate the amount of sand added to the bucket when the system just begins to move, we need to consider the forces acting on the system.

The block has a weight due to gravity, which can be calculated as the mass (28 kg) multiplied by the acceleration due to gravity (9.8 m/s^2).

Weight of the block = mass × acceleration due to gravity
Weight of the block = 28 kg × 9.8 m/s^2
Weight of the block = 274.4 N

Now, we need to consider the static friction between the block and the table. The maximum static friction force is given by the coefficient of static friction (0.45) multiplied by the normal force (which is the weight of the block).

Maximum static friction force = coefficient of static friction × normal force
Maximum static friction force = 0.45 × 274.4 N
Maximum static friction force = 123.48 N

To just begin moving, the force applied to overcome static friction must be equal to the maximum static friction force.

Force applied = 123.48 N

Since the force applied is the weight of the sand added to the bucket, we can calculate the mass of the sand using the formula:

Mass = Force applied / acceleration due to gravity
Mass = 123.48 N / 9.8 m/s^2
Mass = 12.6224 kg

Therefore, the amount of sand added to the bucket is approximately 12.6224 kg.

b) To calculate the acceleration of the system, we need to consider the net force acting on the system. The net force is the difference between the force applied and the force of kinetic friction.

The force of kinetic friction can be calculated as the coefficient of kinetic friction (0.32) multiplied by the normal force (which is the weight of the block).

Force of kinetic friction = coefficient of kinetic friction × normal force
Force of kinetic friction = 0.32 × 274.4 N
Force of kinetic friction = 87.808 N

The net force is given by:

Net force = Force applied - Force of kinetic friction
Net force = 123.48 N - 87.808 N
Net force = 35.672 N

Now, we can calculate the acceleration of the system using Newton's second law of motion, which states that the net force acting on an object is equal to the mass multiplied by the acceleration.

Net force = mass × acceleration

Rearranging the equation, we have:

Acceleration = Net force / mass
Acceleration = 35.672 N / (28 kg + 1 kg)
Acceleration = 1.21 m/s^2

Therefore, the acceleration of the system is approximately 1.21 m/s^2.

1. find time toaccelerate

vf=a t
t= 9/1.5=6 seconds
then find time to deaccelerate
vf=vi+at
0=-vi/a +t t= 9/ 2= 4.5 seconds
now the distance it went at constant velocity
400=1/2 (1.5)*6^2 + 9 tc +9/2 * 4.5
400=27+9tc+10.1
tc=(400-27-10.1)/9=40.3 second
total time: 40.3+4.5+6 seconds.

check all that.